What is the derivative of $f (t) = (- \frac{1}{t + 1}, e^{2 t - 1})$ $f(t) = (-1/(t+1) , e^(2t-1) )$ ?

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What is the derivative of $f (t) = (- \frac{1}{t + 1}, e^{2 t - 1})$ $f(t) = (-1/(t+1) , e^(2t-1) )$ ?

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Answer 1 · 2017-05-07T16:41:28

$\frac{d y}{d x} = {(t + 1)}^{2} e^{2 t - 1}$ $(dy)/(dx)=(t+1)^2e^(2t-1)$

Explanation:

This is a parametric form of equation i.e. $f (t) = (x (t), y (t))$ $f(t)=(x(t),y(t))$

In such a case $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $(dy)/(dx)=((dy)/(dt))/((dx)/(dt))$

Here $x (t) = - \frac{1}{t + 1}$ $x(t)=-1/(t+1)$ and $\frac{d x}{d t} = - \frac{- 1}{{(t + 1)}^{2}} = \frac{1}{{(t + 1)}^{2}}$ $(dx)/(dt)=-(-1)/(t+1)^2=1/(t+1)^2$

and $y (t) = e^{2 t - 1}$ $y(t)=e^(2t-1)$ and $\frac{d y}{d t} = 2 e^{2 t - 1}$ $(dy)/(dt)=2e^(2t-1)$

Hence $\frac{d y}{d x} = \frac{2 e^{2 t - 1}}{\frac{1}{{(t + 1)}^{2}}} = {(t + 1)}^{2} e^{2 t - 1}$ $(dy)/(dx)=(2e^(2t-1))/(1/(t+1)^2)=(t+1)^2e^(2t-1)$

Answer 2 · 2017-05-07T16:42:10

For a parametric function $f (t) = (x (t), y (t))$ $f(t)=(x(t),y(t))$ , the function's derivative is just given by the $x$ $x$ and $y$ $y$ components' derivatives: $f'(t)=(x'(t),y'(t))$ .

So:

$f(t)=(-(t+1)^-1,e^(2t-1))$

$f'(t)=(-(-(t+1)^-2),2e^(2t-1))$

$f'(t)=(1/(t+1)^2,2e^(2t-1))$

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What is the derivative of $f (t) = (- \frac{1}{t + 1}, e^{2 t - 1})$ $f(t) = (-1/(t+1) , e^(2t-1) )$ ?

2 Answers

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